Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int n, cnt, A[MAXN], B[MAXN], pos[MAXN];
vector<pair<int, int> > Ra, Rb;
void lucky() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &B[i]);
}
void Xor(int *A, vector<pair<int, int> > &R, int x, int y) {
  A[x] ^= A[y], R.push_back(pair<int, int>(x, y));
}
void getBase(int *A, vector<pair<int, int> > &R) {
  for (int i = 30, j, p = 1; i >= 0; i--) {
    for (j = p; j <= n && !(A[j] & (1 << i)); j++)
      ;
    if (j > n) continue;
    if (j != p) Xor(A, R, p, j), Xor(A, R, j, p), Xor(A, R, p, j);
    for (j = p + 1; j <= n; j++)
      if (A[j] & (1 << i)) Xor(A, R, j, p);
    p++;
  }
}
int main() {
  lucky();
  getBase(A, Ra), getBase(B, Rb);
  for (int i = 1; cnt++, i <= n && A[i]; i++)
    for (int j = 30; j >= 0; j--)
      if (A[i] & (1 << j)) {
        pos[i] = j;
        break;
      }
  for (int i = 1; i <= cnt; i++)
    for (int j = i; j <= cnt; j++)
      if ((A[i] ^ B[i]) & (1 << pos[j])) Xor(A, Ra, i, j);
  for (int i = 1; i <= n; i++)
    if (A[i] != B[i]) printf("-1\n"), exit(0);
  printf("%d\n", (int)(Ra.size() + Rb.size()));
  for (int i = 0; i < (int)Ra.size(); i++)
    printf("%d %d\n", Ra[i].first, Ra[i].second);
  for (int i = Rb.size() - 1; i >= 0; i--)
    printf("%d %d\n", Rb[i].first, Rb[i].second);
  return 0;
}